#### Learning Objective

- Calculate the pH of a buffer system using the Henderson-Hasselbalch equation.

#### Key Points

- The Henderson-Hasselbalch equation is useful for estimating the pH of a buffer solution and finding the equilibrium pH in an acid-base reaction.
- The formula for the Henderson–Hasselbalch equation is: [latex]pH=p{ K }_{ a }+log(\frac { { [A }^{ - }] }{ [HA] } )[/latex], where pH is the concentration of [H+], pK
_{a}is the acid dissociation constant, and [A^{–}] and [HA] are concentrations of the conjugate base and starting acid. - The equation can be used to determine the amount of acid and conjugate base needed to make a buffer solution of a certain pH.

#### Term

- pKaA quantitative measure of the strength of an acid in solution; a weak acid has a pKa value in the approximate range -2 to 12 in water and a strong acid has a pKa value of less than about -2.

The Henderson–Hasselbalch equation mathematically connects the measurable pH of a solution with the pK_{a }(which is equal to -log K_{a}) of the acid. The equation is also useful for estimating the pH of a buffer solution and finding the equilibrium pH in an acid-base reaction. The equation can be derived from the formula of pK_{a} for a weak acid or buffer. The balanced equation for an acid dissociation is:

[latex]HA\rightleftharpoons { H }^{ + }+{ A }^{ - }[/latex]

The acid dissociation constant is:

[latex]{ K }_{ a }=\frac { [{ H }^{ + }][A^{ - }] }{ [HA] }[/latex]

After taking the log of the entire equation and rearranging it, the result is:

[latex]log({ K }_{ a })=log[{ H }^{ + }]+log(\frac { { [A }^{ - }] }{ [HA] } )[/latex]

This equation can be rewritten as:

[latex]-p{ K }_{ a }=-pH+log(\frac { [A^{ - }] }{ [HA] } )[/latex]

Distributing the negative sign gives the final version of the Henderson-Hasselbalch equation:

[latex]pH=p{ K }_{ a }+log(\frac { { [A }^{ - }] }{ [HA] } )[/latex]

In an alternate application, the equation can be used to determine the amount of acid and conjugate base needed to make a buffer of a certain pH. With a given pH and known pK_{a}, the solution of the Henderson-Hasselbalch equation gives the logarithm of a ratio which can be solved by performing the antilogarithm of pH/pK_{a}:

[latex]{ 10 }^{ pH-p{ K }_{ a } }=\frac { [base] }{ [acid] }[/latex]

An example of how to use the Henderson-Hasselbalch equation to solve for the pH of a buffer solution is as follows:

What is the pH of a buffer solution consisting of 0.0350 M NH_{3} and 0.0500 M NH_{4}^{+ }(K_{a} for NH_{4}^{+} is 5.6 x 10^{-10})? The equation for the reaction is:

[latex]{NH_4^+}\rightleftharpoons { H }^{ + }+{ NH_3}[/latex]

Assuming that the change in concentrations is negligible in order for the system to reach equilibrium, the Henderson-Hasselbalch equation will be:

[latex]pH=p{ K }_{ a }+log(\frac { { [NH_3}] }{ [NH_4^+] } )[/latex]

[latex]pH=9.25+log(\frac{0.0350}{0.0500} )[/latex]

pH = 9.095